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DELTA-CONNECTED LOAD
Load phase currents
V
I AB  AB | I  |  
Z
I BC 
   30   Z
VBC
| I  |    120
Z
VCA
| I  |    120
Z
Line currents
I CA 
Z | Z L |  Z
I aA  I AB  I CA
I bB  I BC  I AB
Method 1: Solve directly
Van | V p | 0
Vbn | V p |   120
Vcn | V p | 120
Positive sequence
phase voltages
Vab  3 | V p | 30
Vbc  3 | V p |   90
Vca  3 | V p |   210
| I line | 3 | I  |
 line     30
Line-phase current
relationship
I cC  I CA  I BC
Method 2: We can also convert the delta
connected load into a Y connected one.
The same formulas derived for resistive
circuits are applicable to impedances
Z
3

|V | / 3
Van
| I aA | AB
I aA 
| I aA |  L  
| Z | / 3
ZY

 L   Z
Balanced case ZY 
| V | 3 | V phase |
    phase  30
Line - phase voltage
relationsh ip
| I line | 3 | I  |
 line     30
Line-phase current
relationship

LEARNING EXTENSION
I aA  1240.
Find the phase currents
I AB  6.9370
I BC  6.93  50
I CA  6.93190
REVIEW OF
Rab  R2 || ( R1  R3 )
Y
Transformations
 Y
Rab  Ra  Rb
Y 
Ra R1
Rb R1
Rb R2
Rb R1
R1 R2


R



R

3
2
Ra
Rc R1
Rc
R2 ( R1  R3 ) Ra  R1  R2  R3 Rb R3
Ra  Rb 
REPLACE IN THE THIRD AND SOLVE FOR R1
R1  R2  R3
R2 R3
Rb 
R1  R2  R3
Ra Rb  Rb Rc  Rc Ra
R3 ( R1  R2 )
R

1
Rb  Rc 
Rb
R
R
3 1
R1  R2  R3 Rc 
R1  R2  R3
R R  Rb Rc  Rc Ra
R2  a b
Rc
 Y
R1 ( R2  R3 )
Rc  Ra 
R R  Rb Rc  Rc Ra
R1  R2  R3
R3  a b
Ra
SUBTRACT THE FIRST TWO THEN ADD
TO THE THIRD TO GET Ra
Y 
R  R1  R2  R3  RY 
R
3
LEARNING EXAMPLE
Delta-connected load consists of 10-Ohm resistance in series
with 20-mH inductance. Source is Y-connected, abc sequence,
120-V rms, 60Hz. Determine all line and phase currents
Van  12030(V )rms
Zinductance  2  60  0.020  7.54
Z  10  j 7.54  12.5237.02  ZY  4.1737.02
VAB 120 360

 16.6022.98( A)rms
Z
10  j 7.54
 16.60  97.02( A)rms
I AB 
I BC
| V | 3 | V phase |
I CA  16.60142.98( A)rms
    phase  30
IaA  28.75  7.02( A)rms
Line - phase voltage
relationsh ip
I bB  28.75  127.02( A)rms
| I line | 3 | I  |
 line     30
Line-phase current
relationship
I cC  28.75112.98( A)rms
Alternatively, determine first the line currents
and then the delta currents
POWER RELATIONSHIPS
| V | 3 | V phase |
- Impedance angle
    phase  30
Line - phase voltage
relationsh ip
Vline
STotal  3 V phase  I *phase
*
STotal  3Vline I line
| I line | 3 | I  |
 line     30
Line-phase current
relationship

Stotal  3Vline  I *
*
STotal  3Vline I line
f

Power factor angle
I line
Ptotal  3 |Vline || I line | cos f
Qtotal  3 |Vline || I line | sin  f
LEARNING EXAMPLE
| Vline | 208(V )rms
Ptotal  1200W
power factor angle  20 lagging

Vline

Determine the magnitude of the line
currents and the value of load impedance
per phase in the delta
- Impedance angle
Ptotal  3 |Vline || I line | cos f
Qtotal  3 |Vline || I line | sin  f
Z  101.4620
Vline
Ptotal | Vline || I line |

cos f | I line | 3.54( A)rms
3
3
| I line | 3 | I  |
 line     30
f

Power factor angle
| Vline |

|
Z
|

 101.46

| I  | 2.05( A)rms
|
I
|
Line-phase current

relationship
I line
LEARNING EXAMPLE
For an abc sequence, balanced Y - Y three phase circuit
source | V phase | 120(V )rms , Zline  1  j1, Z phase  20  j10
Determine real and reactive power per phase at the load and total real, reactive and
complex power at the source
VAN  IaA  (20  j10)  IaA  22.3626.57
VAN  113.15  1.08(V )rms
*
S phase  VAN I aA
 113.15  1.08  5.0627.65
S phase  572.5426.57  512  j 256.09(VA)rms
1200
Chosen
as reference
*
Ssource phase  Van  I aA
 1200  5.0627.65
Van  1200
Vbn  120  120 Because circuit is balanced
Van  120120
data on any one phase is
sufficient
Abc sequence
Van
1200

21  j11 23.7127.65
 5.06  27.65( A)rms
I aA 
Ssource phase  607.227.65
 537.86  j 281.78VA
Ptotal source  3  537.86(W )
Qtotal source  3  281.78 (VA)
Stotal source  Ptotal source  Qtotal source
 1613.6  j845.2(VA)
| Stotal source | 1821.6(VA)
LEARNING EXAMPLE
Determine the line currents and the combined power factor
Circuit is balanced
Load 1 : 24kW at pf  0.6 lagging
Load 2 : 10kW at pf  1

Vline  208(V )rms
Load 3 : 12kVA at pf  0.8 leading

inductive
S  P  jQ
P | S | cos f
P1  24kW

 | S1 | 40kVA
pf  0.6 lagging 
| Q1 | | S1 |2  | P1 |2  32kVA
Q | S | sin  f
pf  cos f
Stotal  S1  S2  S3
f
lagging  inductive  S1  24  j32 kVA
capacitive
Load 2
STOTAL  S1  S2  S3  43.6  j 24.8kVA  50.16029.63kVA
P2  10kW 
  S2  10  j 0 kVA Ptotal  3 |Vline || I line | cos f
| Stotal | 3 | Vline |  | I line |
pf  1


Q

3
|
V
||
I
|
sin

total
line
line
f
Load 3
 f  29.63
| S3 | 12kVA   P3  9.6kW
 
pf  0.8
 | Q3 | 7.2kVA
leading pf  capacitive S3  9.6  j 7.2kVA
pf  0.869 lagging
| I line | 139.23( A)rms
Continued ...
LEARNING EXAMPLE
continued ….
If the line impedances are Z line  0.05  j 0.02
determine line voltages and power factor at the source
inductive
f
capacitive
| I line | 139.23( A)rms
*
Sline  3  ( Z line I line ) I line
 3  Z line | I line |
2
Sline  2908  j1163(VA)
Sload total  43.6  j 24.8kVA  50.16029.63kVA
Ssource total  46.508  j 25.963  53.26429.17kVA
| Stotal | 3 | Vline |  | I line |

 f  29.17
Vline 
53,264
 220.87(V )rms
3  139.13
pf  cos f  cos(29.17)  0.873 lagging
A Y -Y balanced three-phase circuit has a line voltage of
208-Vrms. The total real power absorbed by the load is 12kW
at pf=0.8 lagging. Determine the per-phase impedance
of the load
LEARNING EXTENSION
| V | 3 | V phase |
Impedance angle
    phase  30
Line - phase voltage
relationsh ip
Vline
STotal  3 V phase  I *phase
S  P  jQ
P | S | cos f
208
| V phase |
 120(V )rms
3
Stotal

I line
Q | S | sin  f
pf  cos f
*
 V phase 
| V phase |2
  3
 3V phase  
*
Z

Z
phase
phase


f
Power factor angle
| Z phase|
3 | V phase |2
| Stotal |
 2.88
pf  0.8  cos f   f  36.87
| Stotal |
Ptotal
 15kVA
pf
Z pahse  2.8836.87
LEARNING EXTENSION
Determine real, reactive and complex power at both load
and source
Source is Delta connected.
Convert to equivalent Y
Analyze one phase
j 0.1
10
*
Sload  3  V AN I aA
*
S source  3  Van I aA
 3
| Van |2
*
Z total
phase
10  j 4
120  30
10.1  j 4.1
10.77
| 120
 118.57(V )rms
10.90
VAN 
| VAN
| V AN |2 3 | 118.57 |2 3 | 118.57 |2 (10  j 4)
 3 *


10  j 4
Z phase
102  42
3 | 120 |2 3 | 120 |2 (10.1  j 4.1)


10.1  j 4.1
(10.1) 2  (4.1) 2
Sload  3  (1,212.0  j 484.8)
S source  3  (1224.1  j 496)
LEARNING EXTENSION
A 480-V rms line feeds two balanced 3-phase loads.
The loads are rated
Load 1: 5kVA at 0.8 pf lagging
Load 2: 10kVA at 0.9 pf lagging.
Determine the magnitude of the line current from the 408-V rms source
| S1 | 5kVA 
S  P  jQ
P | S | cos f
P
 P1  4kW
0.8
Q1  | S1 |2  P12  3.0kVA
Q | S | sin  f
pf lagging  S1  4  j3kVA
| S2 | 10kVA 
Q2  | S2 |
2
P
 P  9kW
0.9
 P22
 4.36kVA
S2  9  j 4.36kVA
Stotal  13  j 7.36kVA
| I lineq |
pf  cos f
Stotal  S1  S2
Ptotal  3 |Vline || I line | cos f
Qtotal  3 |Vline || I line | sin  f
| S total | 3 | Vline || I line |
| Stotal |
14,939

 21.14( A)rms
3 | Vline | 706.68